Fourier series of $[x]-x+\frac{1}{2}$: Rigorous justification.

Question

We have the Fourier series $$ [x]-x+\frac{1}{2} = \sum_{n=1}^\infty \frac{\sin 2 \pi n x }{n \pi }. $$ where $x$ is not an integer.

How is this rigorously justified? Are there any references?

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My thoughts:

1. We suppose $f(x)$ can be written in the form $\sum_{k=1}^\infty c_k \sin 2 \pi k x$ and the series is locally uniformly convergent, as $f$ is $1$-periodic, we have

$$ \int_0^1 f(x) \sin 2 \pi n x = c_n \int_0^1 (\sin 2 \pi n x )^2 \, dx = \frac{c_n}{2} \Rightarrow c_n = \frac{1}{n \pi }$$

Answer 1

$f(x)=\lfloor x\rfloor -x-\frac{1}{2}$ is a $1$-periodic function with a Fourier sine series given by $$\lfloor x\rfloor -x-\frac{1}{2} \stackrel{L^2}{=} \sum_{n\geq 1}\frac{\sin(2\pi nx)}{n\pi}\tag{1} $$ since $2\int_{0}^{1}f(x)\sin(2\pi n x)\,dx = \frac{1}{n\pi}$. On the other hand the RHS of $(1)$ is pointwise convergent for any $x\not\in\mathbb{Z}$ by summation by parts, since $$ \left|\sum_{n=1}^{N}\sin(2\pi n x)\right|=\left|\frac{\sin(\pi N x)\sin(\pi(N+1)x)}{\sin(\pi x)}\right|\leq\left|\cot\frac{\pi x}{2}\right|\tag{2}$$ hence the RHS of $(1)$ is pointwise convergent for any $x\not\in\mathbb{Z}$.
The convergence on $\mathbb{R}\setminus\mathbb{Z}$ is not uniform due to Gibbs phenomenon.
However $(2)$ gives that $\sum_{n\geq 1}\frac{\sin(2\pi nx)}{\pi n}$ is uniformly convergent over any interval of the form $I_a=[a+\varepsilon,a+1-\varepsilon]$ for $a\in\mathbb{Z}$. A uniformly convergent series of continuous functions gives a continuous function, hence over $I_a$ we have $$ \sum_{n\geq 1}\frac{\sin(2\pi nx)}{\pi n} \stackrel{\text{unif.}}{=} g(x) \in C^0(I_a) \tag{3}$$ and by $(1)$ it follows that $\int_{I_a}(f(x)-g(x))^2\,dx = 0$. Since $f(x)\in C^0(I_a)$ it follows that $g(x)\equiv f(x)$, hence the convergence given by $(1)$ is also a pointwise convergence over $\mathbb{R}\setminus\mathbb{Z}$.

This is just the Fourier series of a Bernoulli polynomial.

Answer 2

It's probably easier to recall that:

$$\ln(re^{i\theta})=\ln(r)+i\theta$$

$$-\ln(1-e^{i\theta})=\sum_{n=1}^\infty\frac{e^{in\theta}}n=\sum_{n=1}^\infty\frac{\cos(n\theta)+i\sin(n\theta)}n$$

Now take imaginary parts:

$$\sum_{n=1}^\infty\frac{\sin(n\theta)}n=\Im(-\ln(1-e^{i\theta}))=\dots$$

Assuming $\theta\in(0,2\pi)$. For $\theta\notin(0,2\pi)$, it results in a periodic like function...